Ams 261 - Applied Calculus Iii Cheat Sheet Page 2
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1
2
32
2
z
+ y
= 6x
2
2
z
+ x
= 6x
2
z
= 6x
2
z
= 6y
Answer (a) ,
2
2
2
y
+ z
= ( 6x)
= 6x
(7, π/4, π/6)
(7, π/4, 7 3/2)
( 7 3/2, π/4, 7/2)
(7 3/2, π/4, 7/2)
(7/2, π/4, 7 3/2)
(7/2, π/4, 7 3/2)
Answer (e), Use
and
,
,
,
,
2
2
2
2
2
2
r
= ρ
sin
θ
z = ρ cos θ
r
= 49 sin
(π/6) = 49/4
r
= 49/4
r = 7/2
.
z = 7 cos(π/6) = 7 3/2
5
4
6
r (t)
r(t) = 3 cos
ti + 3 sin t
j + t
k
´
=
4
3
4
5
r
(t) =
15 cos
t sin ti + 12t
cos t
sin tj + 6t
k
´
4
3
4
5
r
(t) = 15 cos
t sin ti
12t
cos t
j + 6t
k
´
=
4
3
4
5
r
(t) =
15 cos
t sin ti
12t
cos t
j + 6t
k
´
=
4
3
4
5
r
(t) =
15 cos
t sin ti + 12t
cos t
j + 6t
k
´
4
3
4
5
r
(t) =
15 cos
ti + 12t
cos t
j + 6t
k
Answer (d),
´
d
5
4
6
r
(t) =
[3 cos
ti + 3 sin t
j + t
k]
dt
4
3
4
5
=
(3)(5) cos
t sin ti + (3)(4)t
cos t
j + 6t
k
=
4
3
4
5
=
15 cos
t sin ti + 12t
cos t
j + 6t
k.
a(t) = 6i + 8j + 4k, v(0) = 0 r(0) = 5j
r(1) = 3i + 9j + 2k
r(1) = 3i + 9j
2k
r(1) = 9i + 9j
2k
r(1) = 3i + 5j
2k
r(1) = 3i + 9j + 7k
Answer (a),
v(t) =
a(t)dt = (6t + c
)i + (8t + c
)j + (4t + c
)k
1
2
3
,
v(0) = c
i + c
j + c
k = 0
c
= c
= c
= 0
1
2
3
1
2
3
,
2
2
2
v(t) = 6ti + 8tj + 4tk
r(t) =
v(t)dt = (3t
+ d
)i + (4t
+ d
)j + (2t
+ d
)k
1
2
3
,
,
r(0) = d
i + d
j + d
k = 5j
d
= d
= 0
d
= 5
1
2
3
1
3
2
,
2
2
2
r(t) = 3t
i + (4t
+ 5)j + 2t
k
r(1) = 3i + 9j + 2k
(0, 0, 2) (0, 3, 0)
(5, 0, 0)
Answer
u = 0 0, 0 3, 2 0 = 0, 3, 2 v = 0 5, 3 0, 0 0 =
5, 3, 0
u v =
6i 10j 15k
is normal to the plane. Use standard plane equation with
and
to obtain
n =
6, 10, 15
P (0, 0, 2)
, or
.
6x
10y
15(z
2) = 0
6x
10y
15z + 30 = 0
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